例如我有一个这样的矩阵:
|1 2 3|
|4 5 6|
|7 8 9|
我需要将其转换为这样的矩阵:
|1 4 7|
|2 5 8|
|3 6 9|
实现这一目标的最佳方式是什么?
例如我有一个这样的矩阵:
|1 2 3|
|4 5 6|
|7 8 9|
我需要将其转换为这样的矩阵:
|1 4 7|
|2 5 8|
|3 6 9|
实现这一目标的最佳方式是什么?
DuckDucking打开了这个由肯。令人惊讶的是,它比Nikita的回答更加简洁和完整。它在map()
.
function transpose(a) {
return Object.keys(a[0]).map(function(c) {
return a.map(function(r) { return r[c]; });
});
}
console.log(transpose([
[1,2,3],
[4,5,6],
[7,8,9]
]));
请参阅文章:在 JavaScript 和 jQuery 中转置数组
function transpose(a) {
// Calculate the width and height of the Array
var w = a.length || 0;
var h = a[0] instanceof Array ? a[0].length : 0;
// In case it is a zero matrix, no transpose routine needed.
if(h === 0 || w === 0) { return []; }
/**
* @var {Number} i Counter
* @var {Number} j Counter
* @var {Array} t Transposed data is stored in this array.
*/
var i, j, t = [];
// Loop through every item in the outer array (height)
for(i=0; i<h; i++) {
// Insert a new row (array)
t[i] = [];
// Loop through every item per item in outer array (width)
for(j=0; j<w; j++) {
// Save transposed data.
t[i][j] = a[j][i];
}
}
return t;
}
console.log(transpose([[1,2,3],[4,5,6],[7,8,9]]));
就像在任何其他语言中一样:
int[][] copy = new int[columns][rows];
for (int i = 0; i < rows; ++i) {
for (int j = 0; j < columns; ++j) {
copy[j][i] = original[i][j];
}
}
你只需要在 JS 中以不同的方式构造二维数组。像这样:
function transpose(original) {
var copy = [];
for (var i = 0; i < original.length; ++i) {
for (var j = 0; j < original[i].length; ++j) {
// skip undefined values to preserve sparse array
if (original[i][j] === undefined) continue;
// create row if it doesn't exist yet
if (copy[j] === undefined) copy[j] = [];
// swap the x and y coords for the copy
copy[j][i] = original[i][j];
}
}
return copy;
}
console.log(transpose([
[1,2,3],
[4,5,6],
[7,8,9]
]));
我没有足够的声誉来发表评论(wtf.),所以我需要将Ken 的更新版本作为单独的答案发布:
function transpose(a) {
return a[0].map(function (_, c) { return a.map(function (r) { return r[c]; }); });
}
使用 ES6 中的箭头函数的 Hobs答案的紧凑版本:
function transpose(matrix) {
return Object.keys(matrix[0])
.map(colNumber => matrix.map(rowNumber => rowNumber[colNumber]));
}